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Asked by charlotte on 10 Feb 2013

I have 3x7 images of 256x256 pixels stored in a cell array, i.e. for each pixel i have 7 x-values, 7 y-values and 7 z-values. I want to find the coefficients for z=k1*x + k2*y + k3*x^2 + k4*y^2 + k5*x*y in a least square sense for each pixel without looping over each pixel. Is there a more efficient way to do this?

Answer by ChristianW on 11 Feb 2013

Edited by ChristianW on 13 Feb 2013

Accepted answer

Referring to your 256 sec calulation time:

Got it to 1 sec and 0.8 with parfor. (on my cpu)

dim = 256; C = mat2cell(randi(255,dim*3,dim*7), dim*ones(1,3), dim*ones(1,7)); tic C0 = cellfun(@(x) reshape(x,1,[]),C,'UniformOutput',false); X = cat(1,C0{1,:}); Y = cat(1,C0{2,:}); Z = cat(1,C0{3,:});

K = cell(size(C{1})); for i = 1:size(X,2) % 1:NumberOfPixelsPerImage K{i} = [X(:,i), Y(:,i), X(:,i).^2, Y(:,i).^2, X(:,i).*Y(:,i)]\Z(:,i); end toc

Answer by Image Analyst on 10 Feb 2013

I don't understand your data layout. So you have a cell array with 3 rows and 7 columns. What is inside each cell? Is each cell a 256 by 256 array of either x, y, or z values, like

{256x256 x1 values}, {256x256 x2 values},....{256x256 x7 values}; {256x256 y1 values}, {256x256 y2 values},....{256x256 y7 values}; {256x256 z1 values}, {256x256 z2 values},....{256x256 z7 values};

charlotte on 11 Feb 2013

The data is stored as you described in your code: {256x256 x1 values}, {256x256 x2 values},....{256x256 x7 values}; {256x256 y1 values}, {256x256 y2 values},....{256x256 y7 values}; {256x256 z1 values}, {256x256 z2 values},....{256x256 z7 values};.

I want to find the least square solution for the equations for each pixel: z1=k1*x1 + k2*y1 + k3*x1^2 + k4*y1^2 +k5*x1*y1 z2=k1*x2 + k2*y2 + k3*x2^2 + k4*y2^2 +k5*x2*y1 ... z7=k1*x7 + k2*y7 + k3*x7^2 + k4*y7^2 +k5*x7*y7

Answer by ChristianW on 11 Feb 2013

Is the overall result just 5 scalar k values?

X = cat(1,C{1,:}); Y = cat(1,C{2,:}); Z = cat(1,C{3,:});

M = [X(:), Y(:), X(:).^2, Y(:).^2, X(:).*Y(:)]; K = M\Z(:); % Z = M*K

Show 4 older comments

Matt J on 11 Feb 2013

Even when correctly formulated as a simultaneous system of equations across all pixels, I'm pessimistic that this will give you better performance than a for-loop. By combining into a single system, you lose simplifying information about how the system can be decomposed into smaller problems. I could be wrong, though.

ChristianW on 11 Feb 2013

I'll give it a second shot. I need some help with the math.

Lets talk about a single pixel only. With 7 xValues in X(:,1), each row one of the 7 pictures (analogously for Y and Z), like this:

X = [pixel1_image1 pixel1_image2 ... pixel1_image7];

With these inputs, does this function solve the equations for that pixel?

function K = fcn(X,Y,Z) M = [X, Y, X.^2, Y.^2, X.*Y]; K = M\Z; % Z = M*K

I need a check for that function or an example input with correct output to varify.

Answer by Teja Muppirala on 12 Feb 2013

Here's an approach using sparse matrices to do it. this works in about 0.3 seconds for me, and gives the coefficients in a 5x65536 matrix K.

It should be noted that using a simple for-loop is much simpler to implement, and still works in about 0.6 seconds if you preallocate properly.

% Making random data dim = 256; C = mat2cell(randi(255,dim*3,dim*7), dim*ones(1,3), dim*ones(1,7));

tic C0 = cellfun(@(x) reshape(x,1,[]),C,'UniformOutput',false); X = cat(1,C0{1,:}); Y = cat(1,C0{2,:}); Z = cat(1,C0{3,:});

M = permute( cat(3,X,Y,X.^2,Y.^2,X.*Y), [1 3 2]);

% Generate the locations of the block-diagonal sparse entries jloc = repmat(1:(dim^2*5),7,1); iloc = bsxfun(@plus, repmat((1:7)',1,5) ,reshape( 7*(0:dim^2-1) , 1, 1, [])); SM = sparse(iloc(:),jloc(:),M(:));

% Do the pseudoinversion K = (SM'*SM) \ (SM'*Z(:)); K = reshape(K,5,[]);

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